What is the change in entropy when 535g of ice initially at

What is the change in entropy when 535g of ice initially at -30.0°C absorbs 1.95X10^5 J of energy as heat?

Solution

We have 1.95X10^5 J of energy to spare it will be sufficient to take all the ice 0 degree Celcius of ice and convert some part of it to water at 0 degree celcius.

So

E = ms dt + m H_melting

195000 = 4.17 * 535 * 30 + 333.5 X

X = 383 g

Hence the change in entropy will be

dS = ms ln(Tf/Ti) + m _melted H_melting /T

dS = 535*4.173 ln (243/273) + 383*333.5/273

dS = 727.3 J /K